Predict the products of electrolysis in each of the following:

(i) An aqueous solution of AgNO₃ with silver electrodes.

(ii) An aqueous solution of AgNO₃ with platinum electrodes.

(iii) A dilute solution of H₂SO₄ with platinum electrodes.

(iv) An aqueous solution of CuCl₂ with platinum electrodes.

Given -

All the ions are in aqueous state.

(i) Reaction in solution:

AgNO_3(s) + aq → Ag ⁺ + NO^{3 -}

H₂O ó H ⁺ + OH ^-

At cathode:

Ag ⁺ _(aq) + e ^- →Ag_(s)

Ag ⁺ ions have lower discharge potential than H ⁺ ions. Hence, Ag ⁺ ions get deposited as Ag in preference to H ⁺ ions.

At anode:

Ag_(s)→ Ag ⁺ _(aq) + e ^-

As Ag anode is attacked by NO^{3 -} ions, Ag of the anode will dissolve to form Ag ⁺ ions in the aqueous solution.

(ii) Reaction in solution:

AgNO_3(s) + aq → Ag ⁺ + NO^{3 -}

H₂O óH ⁺ + OH ^-

At cathode:

2Ag ⁺ _(aq) + 2e ^- →2Ag_(s)

Ag ⁺ ions have lower discharge potential than H ⁺ ions. Hence, Ag ⁺ ions get deposited as Ag in preference to H ⁺ ions.

At anode:

2OH ^- _(aq) → O_2(g) + 2H ⁺ _(aq) + 4e ^-

As anode is not attackable, out of OH ^- and NO^{3 -} ions, OH ^- having lower discharge potential, will be discharged in preference to NO^{3 -} ions. These then decompose to give out O₂.

(iii) Reaction in solution:

H₂SO_4(aq) → 2H ⁺ _(aq) + SO₄^{2 -} _(aq)

At cathode:

2H ⁺ _(aq) + 2e ^- →H_2(g)

At anode:

2OH ^- _(aq) → O_2(g) + 2H ⁺ _(aq) + 4e ^-

∴ H₂ gas is evolved at cathode and O_2(g) is evolved at anode.

(iv) Reaction in solution:

CuCl_2(s) + aq → Cu^{2 +} _(aq) + Cl ^- _(aq)

H₂O óH ⁺ + OH ^-

At cathode:

Cu^{2 +} _(aq) + 2e ^- →Cu_(g)

At anode:

2Cl ^- _(aq) - 2e ^- → Cl_2(g)

∴ Cu will be deposited at cathode and Cl₂ gas will be liberated at anode.